Question

If the equation 2cos square theta+cos theta-6=0

Answer 1

SOLUTION :

$\sf \: 2 { \cos}^{2} \theta + \cos \theta - 6 = 0$

$\sf \implies x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}$

$\sf \implies \: \cos \theta = \frac{ - 1 ± \: \sqrt{ {1} + 24 } }{2 \times 2}$

$\sf \implies \: \cos \theta = \frac{ - 1 ± \: 5 }{4}$

$\sf \implies \boxed {\mathfrak{ \cos \theta \: = 1 \: \: \: \: or \: \: \: \: \cos \theta = - \frac{3}{2} }}$